Mongolian Mathematical Olympiad

Let $a$, $b$, $c$ be sides of the triangle. Then $a+b+c=50$. Without losing generality we may assume that $a\ge b\ge c$.

Now $50=a+b+c\le 3a$ and consequently we get $a\ge 17$. On the other hand $50=a+b+c>2a$ by triangle inequality and we came to conclusion $a<25$.

It is well known fact that a triangle with sides $a\ge b\ge c$ is obtuse iff $a^2>b^2+c^2$. Therefore for any pair $(b,c)$ satisfying the above condition $b-c\ge 2$ implies the pair $(b-1,c+1)$ also satisfies. Note that from $2b\ge b+c=50-a$ follows $a\ge b\ge \frac{50-a}{2}$.

Now for $a=17,18,\dots ,24$ form pairs $(b,c)$ such that $a\ge b\ge \frac{50-a}{2}$ and we shall count obtuse triangles with sides $a\ge b\ge c$.

When $a=24$ there are $(b,c)=(24,2),(23,3),\dots ,(13,13)$ pairs. We get from here 11 pairs: $(23,3),(22,4),\dots ,(13,13)$.

When $a=23$ there are $(b,c)=(23,4),(22,5),\dots ,(14,13)$ pairs. We get from here 9 pairs: $(22,5),(21,6),\dots ,(14,13)$.

When $a=22$ there are $(b,c)=(22,6),(21,7),\dots ,(14,14)$ pairs. We get from here 7 pairs: $(20,8),(19,9),\dots ,(14,14)$.

When $a=21$ there are $(b,c)=(21,8),(20,9),\dots ,(15,14)$ pairs. We get from here 3 pairs: $(17,12),(16,13),(15,14)$.

When $a=20$ there are $(b,c)=(20,10),(19,11),\dots ,(15,15)$ pairs. There are no pairs which satisfy the condition.

When $a=19$ there are $(b,c)=(19,12),(18,13),\dots ,(16,15)$ pairs. There are no pairs which satisfy the condition.

When $a=18$ there are $(b,c)=(18,14),(17,15),(15,16)$ pairs. There are no pairs which satisfy the condition.

When $a=17$ there is only one pair: $(b,c)=(17,16)$. But in this case we get an acute triangle.

Thus total number of triangles which satisfy the given condition is $30$.